F.4 binomial theorem^.^

用戶36732

2010-01-07 2:26 am

(1)
It is given that (x^3 + 1/x)^4 + (x^3 – 1/x)^4 = 2(x^12) + p(x^4) + q/(x^4)
(a) Find the values of p and q.
(b) Hence evaluate [(2√2) + (1/√2)]^4 + [(1/√2) – (2√2)]^4.

(2)
It is given that [x/2 + 2/(x^2)]^5 - [x/2 - 2/(x^2)]^5 = a(x^2) + b/(x^4) + c/(x^10).
(a) Find the values of a, b, c.
(b) Hence evaluate [1 + (1/√2)]^5 + [1 - (1/√2)]^5.

*2題都吾洗計(a)部分, 只須計(b)部分
*解得詳細D

Answer :
1
(a) p=12, q=2
(b) 353/2
2
(a) a=5/4, b=40, c=64(b) 29/2

回答 (1)

用戶2053

2010-01-07 3:06 am

✔ 最佳答案

By 1a) : (x^3 + 1/x)^4 + (x^3 – 1/x)^4 = 2(x^12) + 12(x^4) + 2/(x^4)....*
1b) Sub x = √2 to * ,
[(√2)^3 + 1/√2]^4 + [(√2)^3 - 1/√2]^4
= [(2√2) + (1/√2)]^4 + [- 1/√2 + (√2)^3]^4
= [(2√2) + (1/√2)]^4 + [ 1/√2 - (√2)^3]^4
= [(2√2) + (1/√2)]^4 + [(1/√2) – (2√2)]^4
= 2(√2)^12 + 12(√2)^4 + 2/(√2)^4
= 2(2^6) + 12(2^2) + 2/(2^2)
= 353/2
2)By 2a) :
[x/2 + 2/(x^2)]^5 - [x/2 - 2/(x^2)]^5 = (5/4)(x^2) + 40/(x^4) + 64/(x^10)....*
Sub x = √2 to * :
(√2/2 + 1)^5 - (√2/2 - 1)^5 = 10/4 + 10 + 2
(1 + √2/2)^5 + [1 - (√2/2)]^5 = 29/2
[1 + (1/√2)]^5 + [1 - (1/√2)]^5 = 29/2

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